Question Number 119282 by mnjuly1970 last updated on 23/Oct/20
$$\:\:\:\:\:\:\:\:...\:\:{nice}\:\:{calculus}... \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{evaluate}:: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{I}:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {li}_{\mathrm{2}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right){dx}=?? \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:.{m}.{n}.\mathrm{1970}. \\ $$$$\:\:\:\:\:\:\:\: \\ $$
Answered by mindispower last updated on 23/Oct/20
![by part,(d/dx)li_2 (x)=−((ln(1−x))/x)⇒ =[xli_2 (1−x^2 )]_0 ^1 −∫_0 ^1 ((2x^2 ln(x^2 ))/(1−x^2 ))dx =−∫_0 ^1 ((2(x^2 −1+1)ln(x^2 ))/(1−x^2 )) =2∫_0 ^1 ln(x^2 )−2∫_0 ^1 ((ln(x^2 ))/(1−x^2 ))dx 4[xln(x)−x]_0 ^1 −4∫_0 ^1 Σx^(2k) ln(x) −4+4Σ_(k≥0) (1/((2k+1)^2 ))=−4+4.(3/4)ζ(2)=−4+(π^2 /2)](Q119325.png)
$${by}\:{part},\frac{{d}}{{dx}}{li}_{\mathrm{2}} \left({x}\right)=−\frac{{ln}\left(\mathrm{1}−{x}\right)}{{x}}\Rightarrow \\ $$$$=\left[{xli}_{\mathrm{2}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right)\right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}{x}^{\mathrm{2}} {ln}\left({x}^{\mathrm{2}} \right)}{\mathrm{1}−{x}^{\mathrm{2}} }{dx} \\ $$$$=−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}\left({x}^{\mathrm{2}} −\mathrm{1}+\mathrm{1}\right){ln}\left({x}^{\mathrm{2}} \right)}{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left({x}^{\mathrm{2}} \right)−\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({x}^{\mathrm{2}} \right)}{\mathrm{1}−{x}^{\mathrm{2}} }{dx} \\ $$$$\mathrm{4}\left[{xln}\left({x}\right)−\boldsymbol{{x}}\right]_{\mathrm{0}} ^{\mathrm{1}} −\mathrm{4}\int_{\mathrm{0}} ^{\mathrm{1}} \Sigma{x}^{\mathrm{2}{k}} {ln}\left({x}\right) \\ $$$$−\mathrm{4}+\mathrm{4}\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{2}} }=−\mathrm{4}+\mathrm{4}.\frac{\mathrm{3}}{\mathrm{4}}\zeta\left(\mathrm{2}\right)=−\mathrm{4}+\frac{\pi^{\mathrm{2}} }{\mathrm{2}} \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 23/Oct/20
$${perfect}\:{mr}\:{power}\: \\ $$$$\:{thank}\:{you}... \\ $$
Commented by mindispower last updated on 23/Oct/20
$${withe}\:{pleasur} \\ $$